Question E.4: In the circuit of Fig. E.8, let v = 4 cos(5t - 30°) and i = ......
In the circuit of Fig. E.8, let v = 4 \cos (5t – 30° ) and i = 0.8 \cos 5t A Find v_{1} and v_{2}.
As usual, we convert the circuit in the time-domain to its frequencydomain equivalent.
v = 4 \cos(5t – 30°) → V = 4\underline{/30°} , ω = 5
i = 0.8 \cos 5t → I 8 \underline{/0°}
2 H → jωL = j5 × 2 = j10
20 mF → \frac{1}{jωC} = \frac{1}{j10 Ω × 10^{-3}} = – j10
Thus, the frequency-domain equivalent circuit is shown in Fig. E.9. We now apply nodal analysis to this.
At node 1,
\frac{4 \underline{/ – 30°} – V_{1}}{-j 10 } = \frac{V_{1}}{10} + \frac{V_{1} – V_{2}}{j 10 } → 4\underline{/-30°} = 3.468 – j2
= -jV_{1} + V_{2} (E.4.1)
At node 2,
0.8 = \frac{ V_{2}}{20} + \frac{V_{2} – V_{1}}{j10} → j16 = – 2 V_{1} + (2 + j)V_{2} (E.4.2)
Equations (E.4.1) and (E.4.2) can be cast in matrix form as
\begin{bmatrix} – j & 1 \\ -2 & (2 + j) \end{bmatrix} \begin{bmatrix} V_{1} \\V_{2} \end{bmatrix} = \begin{bmatrix} 3.468 – j2 \\ j 16 \end{bmatrix}
Script File
or AV = B . We use MATLAB to invert A and multiply the inverse by B to get V .
>> A = [-j 1 ; -2 (2 + j)]
A =
0 - 1.0000i 1.000
-2.0000 2.0000 + 1.000 i
>> B = [(3.468 - 2j) 16j].’ %note the dot-transpose
B =
3.4680 - 2.0000i
0 + 16.0000i
>> V = inv(A)*B
V =
4.6055 - 2.4403i
5.9083 + 2.6055i
>> abs(V(1))
ans =
5.2121
>> angle(V(1))*180/pi %converts angle from
radians to degrees
ans =
-27.9175
>> abs(V(2))
ans =
6.4573
>> angle(V(2))*180/pi
ans =
23.7973
Thus,
V_{1} = 4.6055 - j2.4403 = 5.212 \underline{/- 27.92°}
V_{2} = 5.908 + j2.605 = 6.457 \underline{/23.8°}
In the time domain,
v_{1} = 4.605 \cos(5t - 27.92°) V , v_{2} = 6.457 \cos(5t + 23.8°) V